Pre-Algebra Examples

$\frac{2 k^{2} + 3 k - 2}{6 k^{2} - 7 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 1

Factor by grouping.

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Step 1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

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Step 1.1.1

Factor $3$ out of $3 k$ .

$\frac{2 k^{2} + 3 (k) - 2}{6 k^{2} - 7 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 1.1.2

Rewrite $3$ as $- 1$ plus $4$

$\frac{2 k^{2} + (- 1 + 4) k - 2}{6 k^{2} - 7 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 1.1.3

Apply the distributive property.

$\frac{2 k^{2} - 1 k + 4 k - 2}{6 k^{2} - 7 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$\frac{(2 k^{2} - 1 k) + 4 k - 2}{6 k^{2} - 7 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 1.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{k (2 k - 1) + 2 (2 k - 1)}{6 k^{2} - 7 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 1.3

Factor the polynomial by factoring out the greatest common factor, $2 k - 1$ .

$\frac{(2 k - 1) (k + 2)}{6 k^{2} - 7 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 2

Factor by grouping.

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Step 2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 6 \cdot 2 = 12$ and whose sum is $b = - 7$ .

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Step 2.1.1

Factor $- 7$ out of $- 7 k$ .

$\frac{(2 k - 1) (k + 2)}{6 k^{2} - 7 (k) + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 2.1.2

Rewrite $- 7$ as $- 3$ plus $- 4$

$\frac{(2 k - 1) (k + 2)}{6 k^{2} + (- 3 - 4) k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 2.1.3

Apply the distributive property.

$\frac{(2 k - 1) (k + 2)}{6 k^{2} - 3 k - 4 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 2.2

Factor out the greatest common factor from each group.

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Step 2.2.1

Group the first two terms and the last two terms.

$\frac{(2 k - 1) (k + 2)}{(6 k^{2} - 3 k) - 4 k + 2} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 2.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{(2 k - 1) (k + 2)}{3 k (2 k - 1) - 2 (2 k - 1)} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $2 k - 1$ .

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{4 k^{2} - 5 k + 1}{k^{2} + k - 2}$

Step 3

Factor by grouping.

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Step 3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 1 = 4$ and whose sum is $b = - 5$ .

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Step 3.1.1

Factor $- 5$ out of $- 5 k$ .

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{4 k^{2} - 5 (k) + 1}{k^{2} + k - 2}$

Step 3.1.2

Rewrite $- 5$ as $- 1$ plus $- 4$

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{4 k^{2} + (- 1 - 4) k + 1}{k^{2} + k - 2}$

Step 3.1.3

Apply the distributive property.

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{4 k^{2} - 1 k - 4 k + 1}{k^{2} + k - 2}$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{(4 k^{2} - 1 k) - 4 k + 1}{k^{2} + k - 2}$

Step 3.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{k (4 k - 1) - (4 k - 1)}{k^{2} + k - 2}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $4 k - 1$ .

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{(4 k - 1) (k - 1)}{k^{2} + k - 2}$

Step 4

Factor $k^{2} + k - 2$ using the AC method.

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Step 4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 4.2

Write the factored form using these integers.

$\frac{(2 k - 1) (k + 2)}{(2 k - 1) (3 k - 2)} \cdot \frac{(4 k - 1) (k - 1)}{(k - 1) (k + 2)}$

Step 5

Simplify terms.

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Step 5.1

Cancel the common factor of $k + 2$ .

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Step 5.1.1

Factor $k + 2$ out of $(2 k - 1) (k + 2)$ .

$\frac{(k + 2) (2 k - 1)}{(2 k - 1) (3 k - 2)} \cdot \frac{(4 k - 1) (k - 1)}{(k - 1) (k + 2)}$

Step 5.1.2

Factor $k + 2$ out of $(k - 1) (k + 2)$ .

$\frac{(k + 2) (2 k - 1)}{(2 k - 1) (3 k - 2)} \cdot \frac{(4 k - 1) (k - 1)}{(k + 2) (k - 1)}$

Step 5.1.3

Cancel the common factor.

$\frac{(k + 2) (2 k - 1)}{(2 k - 1) (3 k - 2)} \cdot \frac{(4 k - 1) (k - 1)}{(k + 2) (k - 1)}$

Step 5.1.4

Rewrite the expression.

$\frac{2 k - 1}{(2 k - 1) (3 k - 2)} \cdot \frac{(4 k - 1) (k - 1)}{k - 1}$

Step 5.2

Multiply $\frac{2 k - 1}{(2 k - 1) (3 k - 2)}$ by $\frac{(4 k - 1) (k - 1)}{k - 1}$ .

$\frac{(2 k - 1) ((4 k - 1) (k - 1))}{(2 k - 1) (3 k - 2) (k - 1)}$

Step 5.3

Cancel the common factor of $2 k - 1$ .

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Step 5.3.1

Cancel the common factor.

$\frac{(2 k - 1) ((4 k - 1) (k - 1))}{(2 k - 1) (3 k - 2) (k - 1)}$

Step 5.3.2

Rewrite the expression.

$\frac{(4 k - 1) (k - 1)}{(3 k - 2) (k - 1)}$

Step 5.4

Cancel the common factor of $k - 1$ .

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Step 5.4.1

Cancel the common factor.

$\frac{(4 k - 1) (k - 1)}{(3 k - 2) (k - 1)}$

Step 5.4.2

Rewrite the expression.

$\frac{4 k - 1}{3 k - 2}$